How Cheenta works to ensure student success?
Explore the Back-Story
Problems and Solutions from CMI Entrance 2022.  Learn More 

Limit of a function | IIT JAM 2017 | Problem 8

Try this problem from IIT JAM 2017 exam. It deals with evaluating Limit of a function.

Limit of a Function | IIT JAM 2017 | Problem 8


Let $$ f(x)=\frac{x+|x|(1+x)}{x} \sin \left(\frac{1}{x}\right), \quad x \neq 0 $$
Write $L=\displaystyle\lim_{x \to 0^{-}} f(x)$ and $R=\displaystyle\lim_{x \to 0^{+}} f(x) .$

Then which one of the following is true?

  • $L$ exists but $R$ does not exist
  • $L$ does not exist but $R$ exists
  • Both $L$ and $R$ exist
  • Neither $L$ nor $R$ exists

Key Concepts


Real Analysis

Function

Limit

Check the Answer


Answer: $L$ exists but $R$ does not exists

IIT JAM 2017 , Problem 8

Try with Hints


Given that, $ f(x)=\frac{x+|x|(1+x)}{x} \sin \left(\frac{1}{x}\right), \quad x \neq 0 $

therefore,

$ f(x)=1+\frac{|x|}{x}(1+x) \sin \left(\frac{1}{x}\right), \quad x \neq 0 $

$ f(x)=\bigg\{\begin{array}{cc}
(2+x) \sin \left(\frac{1}{x}\right), & , x>0 \\
-x \sin \left(\frac{1}{x}\right), & x<0 \\
\end{array} $

Let, $L=\displaystyle\lim_{x \to 0^{-}} f(x)$

and , $R= \displaystyle \lim_{x \rightarrow 0^{+}} f(x) .$

Now,

$L= \displaystyle\lim_{x \to 0^{-}} f(x) $

$\quad = \displaystyle\lim_{x \to 0^{-}} -x \sin \left(\frac{1}{x}\right) $

$ \quad = -\displaystyle\lim_{x \to 0^{-}} x \sin \left(\frac{1}{x}\right) $

Theorem : If $D \subset \mathbb R$ and $f,g : D \to \mathbb R$ . Let $c \in D$. If f is bounded on $N'(c)\cup D$ and $\displaystyle\lim_{x \to c} g(x)=0$, then $\displaystyle\lim_{x \to c}(f.g)(x)=0$.

Now , $ \sin \left(\frac{1}{x}\right) $ is bounded in $\mathbb R - \{0\}$ and $ \displaystyle\lim_{x \to 0^{-}} x=0$ , then $\displaystyle\lim_{x \to 0^{-}} f(x)$ exists and equal to $0$.

But,

$R=\displaystyle\lim_{x\to 0^{+}}f(x)$

$\quad = \displaystyle\lim_{x\to 0^{+}} (2+x) \sin \left(\frac{1}{x}\right),$

$\quad= \displaystyle\lim_{x\to 0^{+}} 2\sin \left(\frac{1}{x}\right) + x \sin \left(\frac{1}{x}\right) $

$\lim_{x \to 0^{+}} \sin \left(\frac{1}{x}\right) $ does not exists [Why?]

Then $L$ exists but $R$ does not.

Subscribe to Cheenta at Youtube


Try this problem from IIT JAM 2017 exam. It deals with evaluating Limit of a function.

Limit of a Function | IIT JAM 2017 | Problem 8


Let $$ f(x)=\frac{x+|x|(1+x)}{x} \sin \left(\frac{1}{x}\right), \quad x \neq 0 $$
Write $L=\displaystyle\lim_{x \to 0^{-}} f(x)$ and $R=\displaystyle\lim_{x \to 0^{+}} f(x) .$

Then which one of the following is true?

  • $L$ exists but $R$ does not exist
  • $L$ does not exist but $R$ exists
  • Both $L$ and $R$ exist
  • Neither $L$ nor $R$ exists

Key Concepts


Real Analysis

Function

Limit

Check the Answer


Answer: $L$ exists but $R$ does not exists

IIT JAM 2017 , Problem 8

Try with Hints


Given that, $ f(x)=\frac{x+|x|(1+x)}{x} \sin \left(\frac{1}{x}\right), \quad x \neq 0 $

therefore,

$ f(x)=1+\frac{|x|}{x}(1+x) \sin \left(\frac{1}{x}\right), \quad x \neq 0 $

$ f(x)=\bigg\{\begin{array}{cc}
(2+x) \sin \left(\frac{1}{x}\right), & , x>0 \\
-x \sin \left(\frac{1}{x}\right), & x<0 \\
\end{array} $

Let, $L=\displaystyle\lim_{x \to 0^{-}} f(x)$

and , $R= \displaystyle \lim_{x \rightarrow 0^{+}} f(x) .$

Now,

$L= \displaystyle\lim_{x \to 0^{-}} f(x) $

$\quad = \displaystyle\lim_{x \to 0^{-}} -x \sin \left(\frac{1}{x}\right) $

$ \quad = -\displaystyle\lim_{x \to 0^{-}} x \sin \left(\frac{1}{x}\right) $

Theorem : If $D \subset \mathbb R$ and $f,g : D \to \mathbb R$ . Let $c \in D$. If f is bounded on $N'(c)\cup D$ and $\displaystyle\lim_{x \to c} g(x)=0$, then $\displaystyle\lim_{x \to c}(f.g)(x)=0$.

Now , $ \sin \left(\frac{1}{x}\right) $ is bounded in $\mathbb R - \{0\}$ and $ \displaystyle\lim_{x \to 0^{-}} x=0$ , then $\displaystyle\lim_{x \to 0^{-}} f(x)$ exists and equal to $0$.

But,

$R=\displaystyle\lim_{x\to 0^{+}}f(x)$

$\quad = \displaystyle\lim_{x\to 0^{+}} (2+x) \sin \left(\frac{1}{x}\right),$

$\quad= \displaystyle\lim_{x\to 0^{+}} 2\sin \left(\frac{1}{x}\right) + x \sin \left(\frac{1}{x}\right) $

$\lim_{x \to 0^{+}} \sin \left(\frac{1}{x}\right) $ does not exists [Why?]

Then $L$ exists but $R$ does not.

Subscribe to Cheenta at Youtube


Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy
Cheenta

Cheenta Academy

Aditya Birla Education Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com
magic-wand